Question

Compare the following pair of surds
$\sqrt{25+36},\sqrt{25}+\sqrt{36}$

Answer 1

$\begin{array}{c}\sqrt{25+36}=\sqrt{61}\\ \sqrt{25}+\sqrt{36}=5+6=11=\sqrt{121}\\ \sqrt{121}>\sqrt{61}\\ \sqrt{25}+\sqrt{36}>\sqrt{25+36}\end{array}$